I am reading the text "An invitation to Algebraic geometry" by Karen Smith et. al. In the text there is a proposition on Veronese mappings:
Prop: The Veronese mapping $v_d$ defines an isomorphism of $\mathbb{P}^n$ onto its image.
In the proof, it is explicitly shown how $\mathbb{P}^n \rightarrow v_d(\mathbb{P}^n) \rightarrow \mathbb{P}^n$ is an identity map, but the other part of the proof is left for the reader. I am struggling to prove that $v_d(\mathbb{P}^n) \rightarrow \mathbb{P}^n \rightarrow v_d(\mathbb{P}^n)$ is an identity map too. Could someone please help me with this?
I am a beginner at Algebraic Geometry, so a detailed explanation would be appreciated. Thanks in advance!
Here is an "elementary" answer. Start with the projective line:
$$v:=v_2: \mathbb{P}^1:=C \rightarrow \mathbb{P}^2:=S$$
defined by
$$v(u:v):=(u^2: uv: v^2).$$
Let $S$ have coordinates $x,y,z$ and prove that $v(C) \cong Z(xz-y^2)$ is the zero set of the polynomial $F:=xz-y^2$. Let $D(x) \subseteq S$ denote the set of "points" in $S$ with non-zero $x$-coordinate (similar for $D(y),D(z)$). You may define maps
$$v_x: D(x) \rightarrow C$$
by
$$v_x(a:b:c):=(a:b).$$
Since $(a:b:c)\in D(x)$ it follows $a \neq 0$ and hence $(a:b)\in C$ is well defined.
$$v_y: D(y) \rightarrow C$$
by
$$v_y(a:b:c):=(a:b) \text{ or }(b:c)$$
Since $b^2=ac$ we have $(a,c \neq 0)$ and it follows
$$(a:b)\cong (ac:bc) \cong (b^2:bc) \cong (b:c)$$
hence $v_y$ is a well defined point in $C$.
$$v_z: D(z) \rightarrow C$$
by
$$v_z(a:b:c):=(b:c)$$
Since $(a:b:c)\in D(z)$ it follows $c \neq 0$ and hence $(b:c)\in C$ is a point.
Then you may verify that $v_x,v_y,v_z$ are well defined maps and local inverses to the map $v_2$, hence $C \cong v(C):=S$ is an isomorphism of algebraic varieties.
Hence this exercise gives you an opportunity to work with local coordinates and to understand what it means for a morphism of algebraic varieties to have an inverse morphism. This method generalize to arbitrary Veronese embeddings and you should do this as an exercise.
Question: "In the proof, it is explicitly shown how $P^n→v_d(P^n)→P^n$ is an identity map, but the other part of the proof is left for the reader. I am struggling to prove that $v_d(P^n)→P^n→v_d(P^n)$ is an identity map too. Could someone please help me with this?"
Answer: In the above example I give an open cover $D(x),D(y),D(z)$ of the variety $v(C)$ and I construct morphisms
$v_x,v_y,v_z$ from $D(x) \rightarrow C$ (resp. $D(y),D(z)$) glueing to a global morphism
$$u: v(C) \rightarrow C$$
with the property that $v \circ u= u \circ v = id$ is the identity. You should verify that the morphisms I define agree on overlaps. It follows $v: C \rightarrow v(C)$ is an isomorphism of varieties.
Note. If $T:=k[x,y,z]/(xz-y^2)$ and $v(C):=Proj(T)$, we have above constructed an isomorphism of varieties
$$ Proj(k[x_0,x_1]):=C \cong^v v(C):=Proj(T),$$
and we may ask for an "inverse ring isomorphism"
$$ f:T \rightarrow k[x_0,x_1]$$
inducing the map $v$ - which is an isomorphism of varieties. You will find there is a natural map $f$ defined by $f(x):=x_0^2, f(y):=x_0x_1, f(z):=x_1^2$ inducing an isomorphism
$$f: k[x,y,z]/(xz-y^2) \cong k[x_0^2,x_0x_1,x_1^2] \subseteq k[x_0,x_1].$$
Hence the "natural" map $f: T \rightarrow k[x_0,x_1]$ is not an isomorphism of $k$-algebras. The correspondence between maps of varieties and maps of coordinate rings is more complicated when studying projective varieties than in the case of affine varieties. In the case of affine varieties $V,W$ with coordinate rings $A(V), A(W)$, there is a 1-1 correspondence between maps $\phi: V \rightarrow W$ and maps of coordinate rings $\phi^*: A(W) \rightarrow A(V)$. Hence there is an isomorphism $V \cong W$ of algebraic varieties (over $k$) iff there is an isomorphism $A(W) \cong A(V)$ of $k$-algebras. This result does not hold for projective varieties.
I'm not familiar with Smith's book, but you also find this explained in Hartshornes book, Chapter II.5, where a relation between maps of graded rings and the Proj construction is outlined.
Note. If you check Exercise II.2.14 in HH you will find a construction of a morphism of schemes
$$f:U \rightarrow Proj(S)$$
where $U \subseteq Proj(T)$ and $\phi: S \rightarrow T$ is a graded morphism of graded rings. In the case of the projective line $C$ there is for every $d\geq 2$ a "natural" sub ring
$$ i:T:=k[x_0^{d-j}x_1^j:j\in\{0,1,..,d\}] \subseteq k[x_0,x_1]$$
and I believe that in the case of $i$ you get an isomorphism
$$\phi_d: C \rightarrow \mathbb{P}^d$$
onto a curve $\phi_d(C)$ - the $d$-uple embedding of $C$. The construction of the Segre embedding and $d$-uple embedding is explained in the exercices in HH.
Note: There is an alternative approach using invertible sheaves. Let $V^*:=k\{y_0,y_1,y_2\}$ and let $S:=\mathbb{P}(V^*)$ and let $\pi: C \rightarrow Spec(k)$ be the structure map. There is a surjective map of sheaves
$$\phi_v: \pi^*(V^*) \rightarrow \mathcal{O}_C(2) \rightarrow 0$$ defined by
$$\phi_v(y_0):=x_0^2, \phi_v(y_1):=x_0x_1, \phi_v(y_2):=x_1^2.$$
By the universal property of $\mathbb{P}(V^*)$ you get a map $v: C \rightarrow \mathbb{P}(V^*)$ and it follows from HH.II.7.2 that $v$ is a closed immersion.
At $D(y_0)$ you get the map
$$v_0: k[\frac{y_1}{y_0}, \frac{y_2}{y_0}] \rightarrow k[\frac{x_1}{x_0}]$$
defined by
$$v_0(\frac{y_1}{y_0}):=\frac{x_1}{x_0}, v_0(\frac{y_2}{y_0}):=(\frac{x_1}{x_0})^2$$
and it follows $v_0$ is surjective.